2w^2+12w-16=0

Solution for 2w^2+12w-16=0 equation:

2w^2+12w-16=0

a = 2; b = 12; c = -16;
Δ = b2-4ac
Δ = 122-4·2·(-16)
Δ = 272
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{17}}{2*2}=\frac{-12-4\sqrt{17}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{17}}{2*2}=\frac{-12+4\sqrt{17}}{4}
$